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Talk:Big boowa
Comparison of SGH and FGH SGH isn't so weak. In fact, for any recusive function p(ω) base on ω, we have \(f_{\omega^{p(\omega)}}(n)\approx p(X)\&n\) and \(g_{p(\omega)}(n)\approx p(n)\). So FGH and SGH first meet at LVO, and \(f_{\theta(\Omega^\Omega)}(n)\approx g_{\theta(\Omega^\Omega)}(n)\) ! {hypcos} (talk) 01:30, October 6, 2013 (UTC) No. If we have \(g_{\theta(\Omega^\omega)+1}(3)\), it will solve to \(g_{\theta(\Omega^3)}(3)\)+1. As you are thinking about it, you'll see that there are no ways to get a higher number in the superscript. So: \(f_{\omega^\omega}(n) \approx g_{\vartheta(\Omega^\omega)}(n)\) \(f_{\omega^\omega}(n+1) \approx g_{\vartheta(\Omega^{\omega+1})}(n)\) \(f_{\omega^\omega}(f_2(n)) \approx g_{\vartheta(\Omega^{\omega^2})}(n)\) \(f_{\omega^\omega}(f_\omega(n)) \approx g_{\vartheta(\Omega^{\varphi(\omega,0)})}(n)\) \(f_{\omega^\omega}(f_{\omega+1}(n)) \approx g_{\vartheta(\Omega^{\Gamma_0})}(n)\) \(f_{\omega^\omega}(f_{\omega^\omega}(n)) \approx g_{\vartheta(\Omega^{\vartheta(\Omega^\omega)})}(n)\) \(f_{\omega^\omega}(f_{\omega^\omega}(f_{\omega^\omega}(n))) \approx g_{\vartheta(\Omega^{\vartheta(\Omega^{\vartheta(\Omega^\omega)})})}(n)\) \(f_{\omega^\omega+1}(n) \approx g_{\vartheta(\Omega^\Omega)}(n)\) Wythagoras (talk) 06:53, October 6, 2013 (UTC) Now I compare SGH Ordinals with FGH and BEAF. SGH Ordinal \(\alpha\) means \(g_{\alpha}(n)\). So \(f_{\omega^\omega+1}(n)\) only reach \(g_{\theta(\Omega^\omega+1)}(n)\). In your opinion, what's wrong? {hypcos} (talk) 06:25, October 9, 2013 (UTC) If you look at it like this: \(f_{\omega^n}(k) \approx g_{\vartheta(\Omega^n)}(k)\) \(f_{\omega^{n+1}}(k) \approx g_{\vartheta(\Omega^{n+1})}(k)\) \(f_{\omega^{2n}}(k) \approx g_{\vartheta(\Omega^{2n})}(k)\) ... but, since SGH can't increase the number inside the brackets, the following comparisons must always be true: \(f_{\omega^n}(k) \approx g_{\vartheta(\Omega^\omega)}(k)\) \(f_{\omega^{n+1}}(k) \approx g_{\vartheta(\Omega^{\omega+1})}(k)\) \(f_{\omega^{2n}}(k) \approx g_{\vartheta(\Omega^{\omega2})}(k)\) ... It is also the case by \(f_\omega(n)\) and \(f_{\omega+1}(n)\): \(f_\omega(n) \approx g_{\vartheta(\omega)}(n)\) \(f_\omega(n+1) \approx g_{\vartheta(\omega+1)}(n)\) \(f_\omega(2n) \approx g_{\vartheta(\omega2)}(n)\) \(f_\omega(n \uparrow\uparrow n) \approx g_{\vartheta(\varepsilon_0)}(n)\) \(f_\omega(f_\omega(n)) \approx g_{\vartheta(\vartheta(\omega))}(n)\) \(f_{\omega+1}(n) \approx g_{\vartheta(\Omega)}(n)\) Wythagoras (talk) 15:29, October 15, 2013 (UTC) \(g_{\theta(\Omega^{\omega+1})}(n)\) isn't \(g_{\theta(\Omega^\omega)}(n+1)=f_{\omega^\omega}(n+1)\). It's \(g_{\theta(\Omega^\omega\theta(\Omega^\omega...\theta(\Omega^\omega)))}(n)\)! {hypcos} (talk) 07:06, October 17, 2013 (UTC) Random section break The problem is that \(g_{\theta(\Omega^\omega,2)}(n) \approx f_{\omega^{2n}}(f_{\omega^\omega}(n))\) is incorrect. Since \(\theta(\Omega^\omega,1)\) is the limit of \(\theta(\Omega,\theta(\Omega^\omega)+1)\), \(\theta(\Omega^2,\theta(\Omega^\omega)+1)\), \(\theta(\Omega^3,\theta(\Omega^\omega)+1)\), ... and \(\theta(\Omega^\omega,2)\) is the limit of \(\theta(\Omega,\theta(\Omega^\omega,1)+1)\), \(\theta(\Omega^2,\theta(\Omega^\omega,1)+1)\), \(\theta(\Omega^3,\theta(\Omega^\omega,1)+1)\), ... we could reach \(\theta(\Omega^\omega,2)\) from \(\theta(\Omega^\omega,1)\), using the same way to reach \(\theta(\Omega^\omega,1)\) from \(\theta(\Omega^\omega)\). The revised comparisons are shown below: So there, \(f_{\omega^\omega+1}(n)\) reaches \(g_{\theta(\Omega^\Omega)}(n)\). ☁ I want more ⛅ 06:56, October 18, 2013 (UTC) :Yes, I tought about it like that Wythagoras (talk) 17:39, October 18, 2013 (UTC) ::Can we conclude from this a way to change a function from FGH to SGH? At least for ordinals in CNF. LittlePeng9 (talk) 18:55, October 18, 2013 (UTC) OK. I see. Now I remove my edits of "approximations" from goobol to dossolplex. {hypcos} (talk) 01:06, October 19, 2013 (UTC) I think Big Boowa is something like \(f_{\vartheta(\Omega_{\omega}) +2}(3)\)....,but then again BEAF isn't well defined at this point. Boboris02 (talk) 17:59, December 3, 2016 (UTC) Below the Small Veblen ordinal, I find that \(f_{3}(n)\) is approximately \(g_{\theta(1)}(n)\) \(f_{\omega}(n)\) is approximately \(g_{\theta(\omega)}(n)\) \(f_{\omega+1}(n)\) is approximately \(g_{\theta(\Omega)}(n)\) \(f_{\omega+2}(n)\) is approximately \(g_{\theta(\Omega+1)}(n)\) \(f_{\omega*2}(n)\) is approximately \(g_{\theta(\Omega+\omega)}(n)\) \(f_{\omega*2+1}(n)\) is approximately \(g_{\theta(\Omega*2)}(n)\) \(f_{\omega*3}(n)\) is approximately \(g_{\theta(\Omega*2+\omega)}(n)\) \(f_{\omega*3+1}(n)\) is approximately \(g_{\theta(\Omega*3)}(n)\) \(f_{\omega^2}(n)\) is approximately \(g_{\theta(\Omega*\omega)}(n)\) \(f_{\omega^2+1}(n)\) is approximately \(g_{\theta(\Omega^2)}(n)\) \(f_{\omega^2+\omega}(n)\) is approximately \(g_{\theta(\Omega^2+\omega)}(n)\) \(f_{\omega^2+\omega+1}(n)\) is approximately \(g_{\theta(\Omega^2+\Omega)}(n)\) \(f_{\omega^2+\omega*2}(n)\) is approximately \(g_{\theta(\Omega^2+\Omega+\omega)}(n)\) \(f_{\omega^2+\omega*2+1}(n)\) is approximately \(g_{\theta(\Omega^2+\Omega*2)}(n)\) \(f_{\omega^2*2}(n)\) is approximately \(g_{\theta(\Omega^2+\Omega*\omega)}(n)\) \(f_{\omega^2*2+1}(n)\) is approximately \(g_{\theta(\Omega^2*2)}(n)\) \(f_{\omega^2*3}(n)\) is approximately \(g_{\theta(\Omega^2*2+\Omega*\omega)}(n)\) \(f_{\omega^2*3+1}(n)\) is approximately \(g_{\theta(\Omega^2*3)}(n)\) \(f_{\omega^3}(n)\) is approximately \(g_{\theta(\Omega^2*\omega)}(n)\) \(f_{\omega^3+1}(n)\) is approximately \(g_{\theta(\Omega^3)}(n)\) \(f_{\omega^4}(n)\) is approximately \(g_{\theta(\Omega^3*\omega)}(n)\) \(f_{\omega^4+1}(n)\) is approximately \(g_{\theta(\Omega^4)}(n)\) \(f_{\omega^\omega}(n)\) is approximately \(g_{\theta(\Omega^\omega)}(n)\) By looking at this, if \(f\) is a successor ordinal, then \(g\) replaces all \(\omega\)'s with \(\Omega\)'s and subtracts 1. If \(f\) is a limit ordinal, then \(g\) is the fundamental sequence of \(f\) by replacing all \(\omega\)'s with \(\Omega\)'s and the variable in the fundamental sequence becomes \(\omega\). Then \(f_{\omega^\omega+1}(n)\) would be \(g_{\theta(\Omega^\Omega)}(n)\) and \(f_{\omega^\omega+1}(n)\) corresponds to \(\{n,n,2(1)2\}\).